If k is a positive integer such that k is a multiple of 6 and a factor of 120, how many po… — GMAT

If $k$ is a positive integer such that $k$ is a multiple of $6$ and a factor of $120$ , how many possible values are there for $k$ ?

The correct answer is 6. First, express 120 in prime factored form as

2^3 \times 3^1 \times 5^1

. Any factor of 120 has the form

2^a \times 3^b \times 5^c

where

0 \le a \le 3

0 \le b \le 1

, and

0 \le c \le 1

. To also be a multiple of 6 (

2^1 \times 3^1

k

must contain at least one factor of 2 and at least one factor of 3. This restricts

a

to 1, 2, or 3 (3 possibilities),

b

to 1 (1 possibility), and

c

to 0 or 1 (2 possibilities). The total number of valid values for

k

3 \times 1 \times 2 = 6

Find the prime factorization of 120 and 6

120 = 2^3 \times 3^1 \times 5^1

and

6 = 2^1 \times 3^1

Prime factorization allows us to express divisibility constraints in terms of exponent bounds.

Determine the exponent constraints for

k = 2^a \times 3^b \times 5^c

For

k

to be a factor of 120,

0 \le a \le 3

0 \le b \le 1

, and

0 \le c \le 1

. For

k

to be a multiple of 6,

a \ge 1

and

b \ge 1

. Thus:

1 \le a \le 3

b = 1

0 \le c \le 1

Combining factor and multiple requirements yields the exact set of possible exponents.

Calculate the total number of combinations for

(a, b, c)

Number of choices for

a

is 3 (1, 2, or 3). Number of choices for

b

is 1 (must be 1). Number of choices for

c

is 2 (0 or 1). Total possible values =

3 \times 1 \times 2 = 6

By the fundamental counting principle, multiplying the number of choices for each exponent gives the total number of valid integers

k

Determining the number of factors of an integer that meet specific divisibility constraints using prime factorization.

Question